My attempt:
With this I know that $\int_A pdA$ and I know that $dA=W\Delta z$ and the equation of a circle is $z^2+(\frac{W}{2})^2=R^2$ and solving for W I get $W=2\sqrt{R^2-Z^2}$ so now the integral is $\int (p)2\sqrt{R^2-Z^2}\Delta z$ ; $p=\rho g z$ so in turn I get $\int_0^R 2\rho g z \sqrt{R^2-Z^2}\Delta z$ or $\int_0^R 2\rho g z \sqrt{R^2-Z^2}dz$
I am not sure if $\int_0^R 2\rho g z \sqrt{R^2-Z^2}dz$ is the right equation to solve. I can integrate this pretty easily, but I am not sure if I have set up the problem correctly and don't want to move on just yet.
pretty late. Yes, its correct. probably why you felt hesitant is because you saw other solutions where z in pgz was further worked on to be expressed as subtraction of another variable. In this problem it's made easier by making the radius of semi-circle exactly equal to (depth of water - that is the z standing for in the hydrostatic equation) because r and z starts right together from the top of the surface.
Other questions had the semi-circle submerged 5 units under water, so z takes into account the total column of water it gained above the submerged object and becomes an expression [ (5+R) - r ]. Remember this r is the point where the differential area is taken so it needs to be integrated.